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soldier1979 [14.2K]
3 years ago
5

Question 3 (1 point)

Chemistry
1 answer:
Serhud [2]3 years ago
3 0

Answer:solid

Explanation:

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Taking advantage of their large differences in pKa values, describe how a mixture of phenol and benzoic acid in diethyl ether so
Marat540 [252]

Answer:

By adding bicarbonate.

Explanation:

The pka of the phenol (C₆H₅OH) is 10 and the pka of the benzoic acid (C₆H₅COOH) is 4, which means that the benzoic acid is a stronger acid than phenol, so if we want to separate phenol from benzoic acid in diethyl ether we need to first use a weak base that will react with benzoic acid and not with the phenol:  

C₆H₅-COOH + HCO₃⁻  ⇄  C₆H₅-COO⁻  +  H₂CO₃

C₆H₅-OH + HCO₃⁻  ⇄  no reaction

The reaction of the benzoic acid with bicarbonate will produce the benzoate ion that will be soluble in the aqueous layer, while the phenol will remain dissolved in the organic layer, so we can separate the two of them by the separation of the two immiscible layers.      

Having the two layers separated, the benzoic acid can be recovered from the aqueous layer by adding HCl:

C₆H₅-COO⁻ + HCl  ⇄  C₆H₅-COOH + Cl⁻

<u>This acid will precipitate from the aqueous solution, and the solid can be isolated by filtration</u>.  

The phenol in the organic layer can be dissolved into an aqueous layer by the adding of a strong base like NaOH:

C₆H₅-OH + OH⁻  ⇄  C₆H₅-O⁻ + H₂O

The phenoxide ion soluble in the aqueous layer can be recovered later by the adding of HCl, which will form the original phenol:

C₆H₅-O⁻ + HCl  ⇄  C₆H₅-OH + Cl⁻  

<u>The precipitated phenol can be isolated by filtration. </u>

This way we can separate a mixture of phenol and benzoic acid in diethyl ether solution.  

I hope it helps you!

6 0
3 years ago
An airplane travels 2100 km at 1000km/hE. It encounters a wind and slows to 800 km/h E for the next 1300 km. What is the average
Deffense [45]

Answer:

The average velocity of the airplane for this trip is 1684.21 km/h

Explanation:

Average velocity is the rate of change of displacement with time. That is,

Average velocity = \frac{Displacement }{Change in time} = Δx / Δt = \frac{x2 - x1}{t2 - t1}

Now we will calculate the time taken by the airplane for the first motion before it encounters a wind.

From,

Velocity = \frac{Distance traveled}{Time taken}

Time = \frac{Distance traveled}{Velocity}

Therefore, Time = \frac{2100km }{1000km/h}

Time = 2.1h

This is the time taken before the airplane encounters a wind.

Hence, t1 = 2.1h

Now, For the time taken by the airplane when it encounters a wind

Also from,

Velocity = \frac{Distance traveled}{Time taken}

Time = \frac{Distance traveled}{Velocity}

Therefore, Time = \frac{1300km }{800km/h}

Time = 1.625h

Hence, t2 = 1.625h

Now, to calculate the average velocity

Average velocity = \frac{x2 - x1}{t2 - t1}

x1= 2100, x2= 1300, t1= 2.1h and t2= 1.625h

Hence, Average velocity = \frac{1300 - 2100}{1.625 - 2.1}

Average velocity = 1684.21 km/h

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Which of the following is a mixture
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The standard unit for measuring mass is the blank
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The answer is (kg) Kilogram
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